anonymous
  • anonymous
Find the vertices and foci of the hyperbola with equation x^2/16- y^2/48=1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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bahrom7893
  • bahrom7893
Just a sec buddy, writing it out. Meanwhile, this link might help you learn more: http://www.purplemath.com/modules/hyperbola2.htm
anonymous
  • anonymous
thank you i'll read that info meanwhile!
bahrom7893
  • bahrom7893
Just a note The vertices are (h-a, k) and (h+a, k) The foci are (h-c, k) and (h+c, k)

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Jhannybean
  • Jhannybean
\[\large \frac{x^2}{16}-\frac{y^2}{48}=1\] foci: \[\large c^2 = a^2 - b^2\]\[a^2 = 16 \ \ b^2 = 48 \]\[c^2 = 16 + 48\]\[c=\sqrt{16+48} = \sqrt{ 64} = 8\] Foci = \(\large \pm 8\)
Jhannybean
  • Jhannybean
Sorry, we know the center is at (0,0) and as @bahrom7893 stated, our foci follow the equation (h-c, k) and (h+c, k) So we have -8 and +8, inputting them into this formula, we would have \[(h+c,k) = (0 + 8, 0) = (8,0) \\ (h-c,k) = (0-8,0) = (-8,0)\]
Jhannybean
  • Jhannybean
And now our vertices....
anonymous
  • anonymous
so my final answer is going to be Vertices: (± 4, 0); Foci: (± 8, 0) ???
Jhannybean
  • Jhannybean
Yeah, believe so.
Jhannybean
  • Jhannybean
Thanks you guys.
anonymous
  • anonymous
thank you so much!

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